Difference between revisions of "2018 AIME I Problems/Problem 6"

Revision as of 23:55, 19 April 2018

Problem

Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ .

Solution 1

Let $a=z^{120}$ . This simplifies the problem constraint to $a^6-a \in \mathbb{R}$ . This is true if $Im(a^6)=Im(a)$ . Let $\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\le\theta<2\pi$ . This must be true for $12$ values of $a$ (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time $\sin\theta=\sin{6\theta}$ ). For each of these solutions for $a$ , there are necessarily $120$ solutions for $z$ . Thus, there are $12*120=1440$ solutions for $z$ , yielding an answer of $\boxed{440}$ .

Solution 2

The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to $0$ . Since $|z|=1$ , let $z=\cos \theta + i\sin \theta$ , then we can write the imaginary part of $\Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0$ . Using the sum-to-product formula, we get $\sin\left(720\theta\right)-\sin\left(120\theta\right)=2\cos\left(\frac{720\theta+120\theta}{2}\right)\sin\left(\frac{720\theta-120\theta}{2}\right)=2\cos\left(\frac{840\theta}{2}\right)\sin\left(\frac{600\theta}{2}\right)\implies \cos\left(\frac{840\theta}{2}\right)=0$ or $\sin\left(\frac{600\theta}{2}\right)=0$ . The former yields $840$ solutions, and the latter yields $600$ solutions, giving a total of $840+600=1440$ solution, so our answer is $\boxed{440}$ .

Solution 3

As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use polar form of complex numbers. Let $z = e^{i \theta}$ . We have two cases to consider. Either $z^{6!} = z^{5!}$ , or $z^{6!}$ and $z^{5!}$ are reflections across the imaginary axis. If $z^{6!} = z^{5!}$ , then $e^{6! \theta i} = e^{5! \theta i}$ . Thus, $720 \theta \equiv 120 \theta \mod 2\pi$ or $600\theta \equiv 0 \mod 2\pi$ , giving us 600 solutions. For the second case, $e^{6! \theta i} = e^{(\pi - 5!\theta)i}$ . This means $840 \theta \equiv \pi \mod 2\pi$ , giving us 840 solutions. Our total count is thus $\boxed{1440}$ .

@@ Line 11: / Line 11: @@
 As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use polar form of complex numbers. Let <math>z = e^{i \theta}</math>. We have two cases to consider. Either <math>z^{6!} = z^{5!}</math>, or <math>z^{6!}</math> and <math>z^{5!}</math> are reflections across the imaginary axis.
 If <math>z^{6!} = z^{5!}</math>, then <math>e^{6! \theta i} = e^{5! \theta i}</math>. Thus, <math>720 \theta \equiv 120 \theta \mod 2\pi</math> or <math>600\theta \equiv 0 \mod 2\pi</math>, giving us 600 solutions.
-For the second case, <math>e^{6! \theta i} = e^{(\pi - 5!\theta)i}</math>. This means <math>840 \theta \equiv \pi mod 2\pi</math>, giving us 840 solutions.
+For the second case, <math>e^{6! \theta i} = e^{(\pi - 5!\theta)i}</math>. This means <math>840 \theta \equiv \pi \mod 2\pi</math>, giving us 840 solutions.
 Our total count is thus <math>\boxed{1440}</math>.

2018 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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